Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {
    public static void main(String args[]) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        double[] p = new double[N];
        for (int i = 0 ; i < N; i++) p[i] = scanner.nextDouble();
        double dp[][] = new double[N+1][N];
        dp[0][0] = (1-p[0]);
        dp[1][0] = p[0];
        for (int i = 1; i < N; i++) {
            dp[0][i] = (1-p[i]) * ((i-1>=0)?dp[0][i-1]:1.0);
            for (int j = 1; j <= i+1; j++) {
                dp[j][i] = ((i-1 >=0 && j-1 >= 0)?dp[j - 1][i - 1]:1.0) * p[i];
                dp[j][i] = dp[j][i] + ((i-1 >= 0)? dp[j][i-1]:1.0) * (1-p[i]);
            }
        }
        double result = 0;
        for (int j = N/2+1; j <= N; j++) {
            result += dp[j][N-1];
        }
        System.out.println(result);
    }
}